What is the integral of 3/(x^2+x+1) ?

The integral of 3/(x^2+x+1) is 2sqrt(3)arctan(1/(sqrt(3))(2x+1))+C

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integral of 3/(x^2+x+1)

\int \frac{3}{x ^{2} + x + 1} d x

23 arctan (\frac{1}{3} (2 x + 1)) + C

Solution steps

\int \frac{3}{x ^{2} + x + 1} d x

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= 3 \cdot \int \frac{1}{x ^{2} + x + 1} d x

Complete the square

x^{2} + x + 1 : (x + \frac{1}{2})^{2} + \frac{3}{4}

= 3 \cdot \int \frac{1}{( x + \frac{1}{2} ) ^{2} + \frac{3}{4}} d x

Apply u-substitution

: \int \frac{4}{4 u ^{2} + 3} d u

= 3 \cdot \int \frac{4}{4 u ^{2} + 3} d u

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= 3 \cdot 4 \cdot \int \frac{1}{4 u ^{2} + 3} d u

Apply Integral Substitution

= 3 \cdot 4 \cdot \int \frac{1}{2 3 ( v ^{2} + 1 )} d v

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= 3 \cdot 4 \cdot \frac{1}{2 3} \cdot \int \frac{1}{v ^{2} + 1} d v

Use the common integral:

\int \frac{1}{v ^{2} + 1} d v = arctan (v)

= 3 \cdot 4 \cdot \frac{1}{2 3} arctan (v)

Substitute back

= 3 \cdot 4 \cdot \frac{1}{2 3} arctan (\frac{2}{3} (x + \frac{1}{2}))

Simplify

3 \cdot 4 \cdot \frac{1}{2 3} arctan (\frac{2}{3} (x + \frac{1}{2})) : 23 arctan (\frac{1}{3} (2 x + 1))

= 23 arctan (\frac{1}{3} (2 x + 1))

Add a constant to the solution

= 23 arctan (\frac{1}{3} (2 x + 1)) + C

\frac{d}{d x} ((x^{2} - x)^{2} - 2 (x^{2} - x))

\frac{d}{d x} (x x + 2)

f (x) = x^{4} - 8 x^{3}

\frac{d}{d x} (cos (- 5 ln (- 3 x - 1)))

\int \frac{( 1 + y )}{y} d y

What is the integral of 3/(x^2+x+1) ?
The integral of 3/(x^2+x+1) is 2sqrt(3)arctan(1/(sqrt(3))(2x+1))+C